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N − 2, if R is large enough. Furthermore, since q ≥ |x | ≥ ωq on 2, const (γ + 1) (γ + 1) ≤ κ N2−1 ≤ const . 30) We now claim that pg γ (x) ≤ − const R at any x ∈ R N for which |g (x)| = 1/2, ∇g (x) = 0 and d is attained on 2 . 31) To prove this, take x as requested here above: then, |d (x)| ≤ |d (x) − r + R + y N +1 | + |r − R| + 1 ≤ |r − R| + 3 ≤ 6 . 30), we get that N −2 κ N −1 d (x) + 2 i=1 const κi 2 d (x) ≤ √ R and so, if R is large enough, N −1 i=1 −κi N −2 2 1 − κi 2 d (x) ≤ −2 κi 2 − κ N2−1 2 i=1 .

38) Flatness of Bernoulli jets 33 This implies that, for any a ∈ ∂ Brω (yω ) and any b ∈ |a − b| ≥ const 1, q2 . 36), we get that (Yω , R1 ) ∩ {|x | ≥ ηq} is at distance at least 3C 0 R inside . 40) Also, yωN + r1 ≤ const q 2 /R, therefore (Yω , R1 ) ⊆ xN ≤ const q 2 R . 41) Also, by inspection, The region X ∈ where |x N +1 | < 1/2 and d (x) is realized on 3 is above S(Yω , R1 ) with respect to the e N +1 -direction. 42) We also claim that The region X ∈ S(Yω , R1 ) where d (Yω ,R1 ) (x) is realized at a point z with |z | ≥ ηq is strictly above in the e N +1 -direction.

Hence, d (Y,R) (ξ ) ≤d (Y,R) (ξ + te N ) . 72), we now select the first t ≥ 0 for which S := t (Y˘ , ˘ ∩ {|x | = c∗ q} ∩ {x N = 0} = ∅ . 76) ˘ which is below {x N = 0} with We also denote by S − the portion of t (Y˘ , R) ˘ which is below S . e. 78) is a small positive constant. t. |x | = r 2 − q2 |x N + r 2 − q 2| . 79) (Y, R). Hence, ) < 0. 76) and the fact that r˘ < r , if x ∈ and therefore d ) . 80) S − , then x is also inside < 0. 82) S− and x ∈ S . for x ∈ Also, since y = 0 by its definition, we gather that, if x ∈ S , then |x − y|2 = |y N |2 + |x |2 = r 2 − q 2 + c∗2 q 2 and therefore d (Y,R) (x ) = |x − y| − r = r 2 − q 2 + c∗2 q 2 − r .

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Flatness of Bernoulli jets by Valdinoci E.


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